Radians/Area

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
TSR launches Learn Together! - Our new subscription to help improve your learning 16-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
Sign in to Reply
  1. Julii92's Avatar
    • Adored and Respected Member
    • Posts: 443
    Radians/Area
    Click image for larger version. 

Name:	Coin.jpg 
Views:	17 
Size:	7.6 KB 
ID:	161061

    A coin is made by starting with an equilateral triangle ABC of side 2cm. With centre A an arc of a circle is drawn joining B to C. Similar arcs join C to A and B. Find the area of the coin.

    I assumed the method would be to find the combined areas of the curved outer segments, then add the area of the equilateral triangle.

    If the area of one of these outer segments is \frac{1}{2}r^2(\theta - \sin \theta) , then the combined area is \frac{3}{2}r^2(\theta - \sin \theta). The triangle ABC is equilateral, so \theta = \frac{\pi}{3}.  r=2, so plugging these numbers into the above equation gives me an area of 1.087cm^2. The area of the equilateral triangle is 2cm^2, so the total area of the face of the coin should be 3.087cm^2.

    This is apparantly the wrong answer, and I don't see how I've gone wrong. Help appreciated.

    Could anyone reccomend any software or websites to draw geometric diagrams? I'm using paint, and doing an awful job, as anyone can see.
  2. CharlieBoardman's Avatar
    • Overlord in Training
    • Location: Manchester
    • Posts: 3,316
    Re: Radians/Area
    The way I did it was treat the triangle and one of the curved outer segments as a sector from a circle with radius 2. To find the area of this, I used \frac{1}{2} r^2 \theta. Using 2 as the radius, and \frac{\pi}{3} as \theta, we get an answer of \frac{2 \pi}{3}.
    Take away the area of the triangle to just get the area of the curved outer segment. Multiply this by 3 (there are 3 segments), then add the area of the triangle back on to get the whole area
    Last edited by CharlieBoardman; 03-07-2012 at 12:37.
  3. Julii92's Avatar
    • Adored and Respected Member
    • Posts: 443
    Re: Radians/Area
    (Original post by CharlieBoardman)
    The way I did it was treat the triangle and one of the curved outer segments as a sector from a circle with radius 2. To find the area of this, I used \frac{1}{2} r^2 \theta. Using 2 as the radius, and \frac{\pi}{3} as \theta, we get an answer of \frac{2 \pi}{3}.
    Take away the area of the triangle to just get the area of the curved outer segment. Multiply this by 3 (there are 3 segments), then add the area of the triangle back on to get the whole area
    Thanks, using your method, I get the final answer as  2{\pi}-2\sqrt3, but according to the textbook the correct answer is  4{\pi}-2\sqrt3. I can't see where I've gone wrong, can you?
  4. CharlieBoardman's Avatar
    • Overlord in Training
    • Location: Manchester
    • Posts: 3,316
    Re: Radians/Area
    (Original post by Julii92)
    Thanks, using your method, I get the final answer as  2{\pi}-2\sqrt3, but according to the textbook the correct answer is  4{\pi}-2\sqrt3. I can't see where I've gone wrong, can you?
    Hmm. I actually get the answer of 2\pi-4.

    \frac{2\pi}{3}-2=\frac{2\pi-6}{3}

    \frac{2\pi-6}{3}\times 3=2\pi-6

    2\pi-6 +2=2\pi-4

    EDIT: I see where I have gone wrong. Stupid mistake. The area of the triangle isn't 2. Give me a minute.
    Last edited by CharlieBoardman; 03-07-2012 at 13:48.
  5. Julii92's Avatar
    • Adored and Respected Member
    • Posts: 443
    Re: Radians/Area
    (Original post by CharlieBoardman)
    Hmm. I actually get the answer of 2\pi-4.

    \frac{2\pi}{3}-2=\frac{2\pi-6}{3}

    \frac{2\pi-6}{3}\times 3=2\pi-6

    2\pi-6 +2=2\pi-4
    I think now you're making the same mistake I did which is to assume that the area of the equilateral trianlge is 2cm^2, when it is actually \sqrt3 cm^2
  6. CharlieBoardman's Avatar
    • Overlord in Training
    • Location: Manchester
    • Posts: 3,316
    Re: Radians/Area
    (Original post by Julii92)
    I think now you're making the same mistake I did which is to assume that the area of the equilateral trianlge is 2cm^2, when it is actually \sqrt3 cm^2
    Yes school boy error! I get the same answer as you. I am assuming that the book is incorrect - unless we have made another error.
    Last edited by CharlieBoardman; 03-07-2012 at 13:57.
  7. Julii92's Avatar
    • Adored and Respected Member
    • Posts: 443
    Re: Radians/Area
    (Original post by CharlieBoardman)
    Yes school boy error! I get the same answer as you. I am assuming that the book is incorrect - unless we have made another error.
    I hope you're right!
  8. J10's Avatar
    • Exalted Member
    Re: Radians/Area
    I also get the same answer as you OP, the book's got it wrong I believe
Sign in to Reply
Share this discussion:  
Article updates
Moderators

We have a brilliant team of more than 60 volunteers looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Reputation gems:
The Reputation gems seen here indicate how well reputed the user is, red gem indicate negative reputation and green indicates a good rep.
Post rating score:
These scores show if a post has been positively or negatively rated by our members.